Given the below function;
[tex]f(x)=(-x^2+7)[/tex]The coordinates of three vertices of a rhombus are (-3, 0), (0, 5) and (3, 0). What are the coordinates of the fourth vertex?
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given coordinates
[tex]A(-3,0),B(0,5),C(3,0),D(x,y)[/tex]STEP 2: State the side properties of a rhombus
In a rhombus, all sides are equal. This means that the length of the sides are equal and therefore the distances of the vertices apart will be the same.
And also, Diagonals of rhombus bisect each other. This implies that:
Co-ordinates of mid-points of AC= Co-ordinates of mid-points of BD
STEP 3: Find the distances of the sides
Midpoints of AC will be calculated as:
[tex]\begin{gathered} \mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right) \\ \left(x_1,\:y_1\right)=\left(-3,\:0\right),\:\left(x_2,\:y_2\right)=\left(3,\:0\right) \\ =\left(\frac{3-3}{2},\:\frac{0+0}{2}\right) \\ =\left(0,\:0\right) \end{gathered}[/tex]Midpoints of BD will be calculated as:
[tex]\begin{gathered} \mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right) \\ \left(x_1,\:y_1\right)=\left(0,\:5\right),\:\left(x_2,\:y_2\right)=\left(x,\:y\right) \\ =\left(\frac{x+0}{2},\:\frac{y+5}{2}\right) \\ =\left(\frac{x}{2},\:\frac{y+5}{2}\right) \end{gathered}[/tex]Since midpoints are the same as mentioned above, this means that:
[tex]\begin{gathered} \left(\frac{x}{2},\:\frac{y+5}{2}\right)=(0,0) \\ \frac{x}{2}=0,x=0 \\ \frac{y+5}{2}=0,y+5=0,y=-5 \\ \\ \therefore(x,y)=(0,-5) \end{gathered}[/tex]Hence, the coordinates of the fourth vertex is given as:
[tex][/tex]Find the missing dimension of the figure shown to the right round to the nearest tenth.
On the right triangle, we know the measure of the hypotenuse and one of its sides, then, using the pythagorean theorem, we get:
[tex]x=\sqrt[]{(29)^2-(14)^2}=\sqrt[]{841-196}=\sqrt[]{645}=25.4[/tex]therefore, the missing dimension is 25.4''
I need help with this question can you please help me ?
Answer:
Given that,
The product of a number and 6 equals twice the result of the sum of the number and 6.
Let the number be x,
product of a number and 6 is 6x.
sum of the number and 6 is x+6.
we get,
[tex]6x=2(x+6)[/tex]a) The equation could be used to find the number is,
[tex]6x=2(x+6)[/tex]b) On solving the above equation we get,
[tex]6x=2x+12[/tex][tex]6x-2x=12[/tex][tex]4x=12[/tex][tex]x=3[/tex]The number is 3.
Answer is: 3
What is the lateral surface area of the prism shown below? 9 m 6m 6 m 10 m
we are asked to find the surface area of a prism. To do that, we will find the areas of each lateral rectangle and the areas of the top and bottom triangles.
The areas of the rectangles are:
[tex]\begin{gathered} A_1=(10)(6)=60 \\ A_2=(10)(6)=60 \\ A_3=(10)(50)=50 \end{gathered}[/tex]To determine the ara of the top triangle we will use the following formula:
[tex]undefined[/tex]you are trying to upload a photo for your school picture,but you need to reduce the size of the photo by a quarter of its original size. Which is the correct power that will reduce the picture by a quarter?A: 2^2B: 2^4C: 2^-2D: 2^-4
the scale factor is 1/4, which is equivalent to:
[tex]\frac{1}{4}=\text{ }\frac{1}{2^2}=2^{-2}[/tex]In this diagram, ABAC – AEDF. If thearea of ABAC = 6 in2, what is thearea of AEDF?DAE 2 inB3 inс=Area = [? ] in2Enter a decimal rounded to the tenths.a
Area of ΔBAC = 6 in^2
EF = 2 in
BC = 3 in
Both triangles are similar, so:
Area ΔBAC : Area of ΔEDF = BC^2 : EF^2
Replacing:
6 / Area of ΔEDF = 3^2 / 2^2
Cross multiply
6 * 2^2 = 3^2 * Area of ΔEDF
24 = 9 * Area of ΔEDF
24/9 = Area of ΔEDF
Area of ΔEDF = 8/3 in^2 = 2.7 in^2
Please help with number 8Solve each equation by completing the square.simplify all irrational and complex situations
We're going to solve by completing the square the given equation:
5x²+14x=3 (divide both sides by 5)
x² +(14/5)x=3/
Multiplication Lattice Model) 1 bicycle costs $ 215. 1. Use the lattice model to multiply 2. How much will be paid for 87 bicycles 3. Upload the work
1. Multiply the number on the top of the box with the number to the right of the box. Write the answer in the box (write the product one digit on the top of the green line and the other digit under that line:
Example 2 x 8 = 16 write the 16 as in the image above
2. Add the numbers in the box diagonally and carry to the next column when necessary. Start from right to left
Then, 215 multiplied by 87 is 18705.
For 87 bicycles cost $18705Help me please and thank you I just need part a and d please
a. To complete the table, we have to assign the corresponding study hours per week to the hours worked per week.
For example, for the first interval 0≤h<4, the credits taken are 18 (according to the first table). When the credits taken are 18, the study hours per week are 39 (according to the second table). It means that for the interval 0≤h<4, the study hours per week is 39.
The values of this table will be, respectively:
39, 32, 24, 21, 17, 14, 12.
d. An student works between 12 and 16 hours every week, that allows him take 15 credits, which means that he has to use 21 hours to study. He wants to study 32 hours per week, that way, he can take 17 credits. To achieve it he needs to reduce its working hours to 4 to 8 hours per week, that way he would be able to take 17 credits and study 32 hours per week.
Please I need help finding the equation of the parallel line and the perpendicular line.
The equation parallel to the given equation and passing through the point (8, 3) is:
[tex]y\text{ = }\frac{5}{2}x\text{ - 17}[/tex]The equation perpendicular to the given equation and passing through the point (8, 3) is:
[tex]y\text{ = }\frac{-2}{5}x\text{ + }\frac{31}{5}[/tex]Explanations:The equation of the line parallel to the line y = mx + c and passing through the point (x₁, y₁) is given as:
[tex]y-y_1=m(x-x_1)[/tex]The equation of the line perpendicular to the line y = mx + c and passing through the point (x₁, y₁) is given as:
[tex]y-y_1\text{ = }\frac{-1}{m}(x-x_1)[/tex]Now, for the equation:
[tex]\begin{gathered} y\text{ = }\frac{5}{2}x\text{ - 7} \\ m\text{ = }\frac{5}{2} \end{gathered}[/tex]The line parallel to the equation and passing through the point (8, 3) will be:
[tex]\begin{gathered} y\text{ - 3 = }\frac{5}{2}(x\text{ - 8)} \\ y\text{ - 3 = }\frac{5}{2}x\text{ - 20} \\ y\text{ = }\frac{5}{2}x\text{ - 20 + 3} \\ y\text{ = }\frac{5}{2}x\text{ - 17} \end{gathered}[/tex]The line perpendicular to the given equation and passing through the point (8, 3) will be:
[tex]\begin{gathered} y\text{ - 3 = }\frac{-2}{5}(x\text{ - 8)} \\ y\text{ - 3 = }\frac{-2}{5}x\text{ + }\frac{16}{5} \\ y\text{ = }\frac{-2}{5}x\text{ + }\frac{16}{5}+3 \\ y\text{ = }\frac{-2}{5}x\text{ + }\frac{31}{5} \end{gathered}[/tex]Tell whether the rational number is a reasonable approximation of the square root.1.) 277/160, square root of 3
we know that
[tex]\begin{gathered} \sqrt{3}=1.73205 \\ \frac{277}{160}=1.73125 \end{gathered}[/tex]therefore
the rational number is a reasonable approximation of the square rootConsider the following word problem:Two planes, which are 1180 miles apart, fly toward each other. Their speeds differ by 40 mph. If they pass each other in 2 hours, what isthe speed of each?Step 2 of 2: Solve the equation found in Step 1.
To answer this question, we can state the problem as follows:
1. The two planes are 1180 miles apart.
2. They fly toward each other.
3. Their speed differs by 40 mph: that is one of the planes is faster than the other 40 mph, or the other plane is slower than the other plane.
4. The time they encounter each other is 2 hours.
Then, we need to remember the formula for a constant speed:
[tex]V=\frac{d}{t}[/tex]Where
• d is the distance
,• t is the time
Then, we have that the speeds for each of the planes are:
[tex]V_{p1}=V_{p2}+40[/tex]We also have that the sum of the distance for both planes is 1180 miles:
[tex]d_{p1}+d_{p2}=1180[/tex]And we have that:
[tex]V=\frac{d}{t}\Rightarrow d=V\cdot t[/tex][tex]d_{p1}=V_{p1}\cdot t[/tex][tex]d_{p2}=V_{p2}\cdot t[/tex]But
[tex]V_{p1}=V_{p2}+40[/tex]Then, we have that:
[tex]d_{p1}+d_{p2}=1180[/tex][tex]V_{p1}\cdot t+V_{p2}\cdot t=1180\Rightarrow V_{p1}=V_{p2}+40[/tex][tex](V_{p2}+40)\cdot t+V_{p2}\cdot t=1180[/tex]Since t = 2, we have:
[tex](V_{p2}+40)\cdot2+V_{p2}\cdot2=1180[/tex][tex]2V_{p2}+80+2V_{p2}=1180\Rightarrow4V_{p2}+80=1180[/tex]Subtracting 80 from both sides of the equation:
[tex]4V_{p2}+80-80=1180-80\Rightarrow4V_{p2}=1100[/tex]Dividing both sides of the equation by 4, we have:
[tex]V_{p2}=\frac{1100}{4}\Rightarrow V_{p2}=275\text{mph}[/tex]Since we know that:
[tex]V_{p1}=V_{p2}+40[/tex]Then, we have:
[tex]V_{p1}=275\text{mph}+40\text{mph}\Rightarrow V_{p1}=315\text{mph}[/tex]In summary, therefore, the speed of each plane is:
• Speed of Plane 1 = 315 mph
,• Speed of Plane 2 = 275 mph
Suppose that the dollar value v(t) of a certain car that is t years old is given by the following exponential function.v (t) = 19,900(0.91)^tFind the initial value of the car.Does the function represent growth or decay?By what percent does the value of the car change each year?
The general form of an exponential function is:
[tex]f(x)=ab^x[/tex]where a = initial value and b = the rate of growth.
In the given equation that we have,
[tex]v(t)=19,900(0.91)^t[/tex]we can see that the value of a = 19, 900. Hence, this is the initial value of the function and thus, the initial value of the car is 19, 900.
We can also see that the rate of growth is 0.91. Since the rate of growth is between 0 and 1, the function represents exponential decay.
The value of the car decreases by 9% each year.
Use the figures below. What is the ratio of AD to JM? A.2/3B.6/5C.3/2
Given:
Rectangle ABCD is similar to rectangle JKLM
From the first rectangle, we have:
AB = 15
DC = 6
From the second rectangle, we have:
JM = 10
ML = 5
We know that,
AD ~ JM
DC ~ ML
Thus, we have the ratio as:
[tex]\frac{AD}{JM}=\frac{DC}{ML}[/tex][tex]\begin{gathered} \frac{AD}{JM}=\frac{15}{10}=\frac{3}{2} \\ \end{gathered}[/tex]Therefore the ratio of AD to JM is:
[tex]\frac{3}{2}[/tex]ANSWER:
[tex]C\text{. }\frac{3}{2}[/tex]Find the approximate mean and standard deviation of the following data.
Given the data in the table, to find the approximate mean we would need to find the midpoints for the given intervals.
The midpoint was gotten from the average of the upper and lower class intervals.
We can then find the approximate mean using the formula below.
[tex]\bar{x}(Mean)=\frac{\sum^{}_{}fx}{\sum^{}_{}f}[/tex]
Therefore,
[tex]\begin{gathered} \bar{x}=\frac{(8\times3.5)+(11.5\times3)+(19.5\times9)+(27.5\times5)}{8+3+9+5} \\ \bar{x}=\frac{28+34.5+175.5+137.5}{25} \\ \bar{x}=\frac{375.5}{25} \\ \bar{x}=15.02 \end{gathered}[/tex]To find the standard deviation we would use the formula below;
[tex]\begin{gathered} \sigma^2=\frac{\sum^{}_{}f(x-\bar{x})^2}{\sum^{}_{}f} \\ ^{} \end{gathered}[/tex]Thus;
[tex]\begin{gathered} \sigma^2=\frac{8(3.5-15.02)^2+3(11.5-15.02)^2+9(19.5-15.02)^2+5(27.5-15.02)^2}{25} \\ =\frac{1061.6832+37.1712+180.6336+778.752}{25} \\ \sigma^2=82.3296 \\ \sigma=\sqrt[]{82.3296} \\ \sigma=9.074 \end{gathered}[/tex]Answer:
Therefore, the approximate mean and standard deviation is 15.02 and 9.074 respectively
The figure below was made with a scale of 1 unit = 9 cm.Draw the figure with a new scale of 1 unit = 3 cm.You can place your figure anywhere on the grid on the right.9 cmCurrent scale1 unit = 9 cmExplanationCheck3 cmNew scale1 unit = 3 cmI need help with this math problem
We will draw the figure
In a new scale, the new scale is
[tex]1\text{ unit= 3 cm}[/tex]Note that the draw above is a square of side 18cm, therefore in the new scale the side of the square have to be drawn using
[tex]\frac{18cm}{3cm}=6\text{ }units[/tex]That is, if we change the units our new square have a side of 6 units as follows
Evaluate the expression when b= 6 and x = -5. -b+ 2x
Given the below expression;
[tex]-b+2x[/tex]Evaluating the expression when b = 6 and x = -5, we'll have;
[tex]\begin{gathered} -(6)+2(-5) \\ -6-10 \\ -16 \end{gathered}[/tex]Therefore, when b = 6 and x = -5, the expression will be -16.
2/7:12/7 how to divide
In order to divide two fractions, we cross-multiply them:
This way,
[tex]\frac{2}{7}\div\frac{12}{7}\rightarrow\frac{2\cdot7}{7\cdot12}\rightarrow\frac{14}{84}\rightarrow\frac{2}{12}\rightarrow\frac{1}{6}[/tex]Therefore, we can conclude that
[tex]\frac{2}{7}\div\frac{12}{7}=\frac{1}{6}[/tex]Rewrite the following equation in slope intercept form3x + 17y = -4Write your answer using integers, proper fractions, and improper fractions in simplest form.
the given expression is,
3x + 17y = -4
17y = -4 - 3x
y = -3/17x - 4/17
thus, the answer is,
[tex]y=-\frac{3}{17}x-\frac{4}{17}[/tex]Perform the indicated operation. Write your solution in scientific notation: 0.00000936 0.00000003 a. 3.12 x 102 son b. 3.12 x 10-14 C. 3.12 x 1048 3 d. 3.12 x 104
1) To write a scientific notation number we must write a rational number followed by its power of 10.
2) So we can write:
[tex]undefined[/tex]Find g′(4) given that f(4)=5, f′(4)=−1, and g(x)=(√x)*f(x).
Given that:
[tex]g(x)=\sqrt[]{x}f(x)[/tex]You need to find:
[tex]g^{\prime}(x)[/tex]In order to derivate the function, you need to apply the Product Rule
[tex]\frac{d}{dx}(u\cdot v)=u\cdot v^{\prime}+v\cdot u^{\prime}[/tex]Then, you get:
[tex]g^{\prime}(x)=\sqrt[]{x}\cdot f^{\prime}(x)+f(x)(\sqrt[]{x})^{\prime}[/tex]Since:
[tex]\sqrt[]{x}=x^{\frac{1}{2}}[/tex]You know that:
[tex]\frac{d}{dx}(\sqrt[]{x})=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt[]{x}}[/tex]Hence:
[tex]\begin{gathered} g^{\prime}(x)=\sqrt[]{x}\cdot f^{\prime}(x)+f(x)(\frac{1}{2\sqrt[]{x}}) \\ \\ g^{\prime}(x)=\sqrt[]{x}\cdot f^{\prime}(x)+\frac{1}{2\sqrt[]{x}}f(x) \end{gathered}[/tex]Knowing that you need to find:
[tex]g^{\prime}(4)[/tex]You can rewrite the function as follows:
[tex]g^{\prime}(4)=\sqrt[]{4}\cdot f^{\prime}(4)+\frac{1}{2\sqrt[]{4}}f(4)[/tex]Knowing that:
[tex]\begin{gathered} f\mleft(4\mright)=5 \\ f^{\prime}\mleft(4\mright)=-1 \end{gathered}[/tex]You can substitute values:
[tex]g^{\prime}(4)=(\sqrt[]{4})(-1)+(\frac{1}{2\sqrt[]{4}})(5)[/tex]Evaluating, you get:
[tex]\begin{gathered} g^{\prime}(4)=(2)(-1)+(\frac{1}{2\cdot2})(5) \\ \\ g^{\prime}(4)=-\frac{3}{4} \end{gathered}[/tex]Hence, the answer is:
[tex]g^{\prime}(4)=-\frac{3}{4}[/tex]
What is the product of 13/78 and 78/13?
10. Find f(-3) + 1 using the following equation f(x) = 5x – 4
We are to find the value of f(-3) + 1
using the following expression for f(x):
f(x) = 5 x - 4
Then f(-3) = 5 (-3) - 4 = -15 - 4 = - 19
since we need to add 1 to this result, we get:
f(-3) + 1 = -19 + 1 = - 18
What’s the mid point of AB in the picture below
From the given linear graph, we would write out the co-ordinates of the points A and B first in the form of (x,y).
Thus, we have:
[tex]\begin{gathered} A(-6,-4) \\ B(-3,3) \end{gathered}[/tex]The mid-point of a line segment;
[tex]\begin{gathered} A(x_1,y_1)\text{ and} \\ B(x_2,y_2) \end{gathered}[/tex]is given as:
[tex](\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]Thus, we have:
[tex]\begin{gathered} (\frac{-6+(-3)}{2},\frac{-4+3}{2}) \\ (\frac{-6-3}{2},-\frac{1}{2}) \\ (\frac{-9}{2},-\frac{1}{2}) \\ (-4.5,-0.5) \end{gathered}[/tex]Hence, the midpoint of the line segment AB is: ( -4.5, -0.5)
Tyrone's car averages 26.2 miles per gallon in the city and 12.2 more miles per gallon on the highway. The gas tank holds 15 gallons and is 2/3 of the way full. With his current gas level, approximately how many miles can Tyrone drive on the highway?
First, we need to find the nubser of miles it can rive on high-way
From the question Tyron's car can drive 26.2 + 12.2 = 38.4 miles on the high way
Next, we find gallon of gas level that is currently on his car
His car can hold 15 gallons, but his current gas level is 2/3
So we will find 2/3 of 15 gallon which is equal;
2/3 x 15 = 30/3 = 10 gallons
Now, using proportion;
Let X be the number of miles he can drive with 10 gallons
38.4 miles = 1 gallon
X = 10 gallons
X = 38.4 x 10
X = 384 miles
Therefore he can drive 384 miles on highway
Which answers describe the shape below? Check all that apply.A. SquareB. RhombusC. QuadrilateralD. TrapezoidE. RectangleF. Parallelogram
Recall the following definitions:
A square is a shape that has 4 sides. All of them with the same length. It has 2 pairs of parallel sides and has 4 right angles. Based on this definition, the given shape is a square.
A rhombus is a shape that has 4 sides. It has two pairs of parallel sides and each par has the same length. Based on this definition, it is also a rhombus.
A quadrilateral is closed shape that has 4 sides. Based on this definition, this shape is also a quadrilateral.
A trapezoid is a quadrilateral that has exactly one pair of parallel sides. As in this case we have two pairs of parallel sides, it is not a trapezoid.
A rectangle is a quadrilateral that has a pair of parallel sides of equal length and has 4 right angles. Based on this definition, this shape is also a rectangle.
A parallelogram is a quadrilateral that has two pairs of parallel sides with the same length. Based on this definition, this shape is also a parallelogram
what word represents when the line intersects with the y-axis
Given:
what word represents when the line intersects with the y-axis?
The intersection between a line and the y-axis gives:
the point of the y-intercept
A house is worth $350,000 when purchased. It was worth $335,000 after the firstyear and $320,000 after the 2ndyear.1. Geometric or Arithmetic and Why?2. Complete a table that shows the value of the house for 5 years.3. Write an explicit and recursive formula for the sequence.4. What is the value of the house after you have lived in it for 10 years?
A house is worth $350,000 when purchased. It was worth $335,000 after the first year and $320,000 after the 2nd year.
So, the difference between initial cost and the cost after one year =
335,000 - 350,000 = -15,000
The difference between the cost after one year and after 2 years =
320,000 - 335,000 = -15,000
As the common difference is constant
so, the cost represents Arithmetic sequence
the first term is 330,000 and the common difference is -15,000
The general form of the sequence is a + d(n - 1)
where a is the first term and d is the common difference and n the number of term
so, a = 335,000 and d = -15,000
so, the general form will be = 335,000 - 15,000(n-1)
So, the value of the house after 5 years = 335,000 - 15,000 * (5-1) = 275,000
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1. Geometric or Arithmetic and Why?
Arithmetic
2. Complete a table that shows the value of the house for 5 years.
For 5 years:
first year = $335,000
second year = $320,000
third year = $305,000
fourth year = $290,000
fifth year = $275,000
3. Write an explicit and recursive formula for the sequence.
The formula will be : 335,000 - 15,000(n-1)
4. What is the value of the house after you have lived in it for 10 years?
After 10 years;
the value of the house = 335,000 - 15,000 * (10-1)
= 335,000 - 15,000 * 9 = $200,000
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I need further explanation on number 3 please enlighten me will give 5 stars if you give me a good explanation ⭐️⭐️⭐️⭐️⭐️
To evaluate an expression at a given value means that you substitute the given value anywhere you see the variable. For example, if the expression is:
[tex]x+1[/tex]and the problem asks you to evaluate the expression at x=9, then we substitute x=9 and get:
[tex]9+1=10.[/tex]Now, question number 3 asks us to evaluate
[tex](1.42+t)\times u+\frac{0.42}{v},[/tex]at t=13.5, u=7, and v=6. Therefore substituting the given values we get:
[tex]\begin{gathered} (1.42+13.5)\times7+\frac{0.42}{6}=(14.92)\times7+0.07 \\ =104.44+0.07=104.51. \end{gathered}[/tex]Answer: 104.51
option 2 you have 1 hour of homework at the start, then amount of homework duobles every week.
As per given by the question,
There are given that to the construct of equation to reperesent the number of hours of home work.
Now,
Here, y for the total hours and x for the total weeks.
The,
From option two;
In starting, you have 1 hours of homework, then after that amount of homework double in every week.
So,
Total hours of homework is 1,
Then , y is represent the 1 hour of homework.
Now,
The amount of home work double in every week,
So,
[tex]2y[/tex]Here, x represent the week.
So, homework double in every week.
Hence, the equation is;
[tex]x=2y[/tex]